Sea water is found to contain `5.85% NaCI` and`9.50% MgCI_(2)` by weight of solution. Calculate its normal boiling point assuming `80%` ionisation for `NaCI` and `50%` ionisation of `MgCI[K_(b)(H_(2)O) =0.51 kg mol^(-1)K]`

Explanation

To solve the problem of calculating the normal boiling point of seawater containing 5.85% NaCl and 9.50% MgCl₂, we will follow these steps:

### Step 1: Calculate the weight of the solvent (water)
Given that the total weight of the solution is 100 g, we can find the weight of the solvent (water) by subtracting the weights of NaCl and MgCl₂ from the total weight.

\[ \text{Weight of NaCl} = 5.85 \, \text{g} \]
\[ \text{Weight of MgCl}_2 = 9.50 \, \text{g} \]
\[ \text{Weight of solution} = 100 \, \text{g} \]
\[ \text{Weight of solvent (water)} = 100 \, \text{g} - (5.85 \, \text{g} + 9.50 \, \text{g}) = 100 \, \text{g} - 15.35 \, \text{g} = 84.65 \, \text{g} \]

### Step 2: Calculate the moles of NaCl and MgCl₂
Next, we will calculate the number of moles of NaCl and MgCl₂ using their respective molar masses.

- **Molar mass of NaCl** = 58.5 g/mol
- **Molar mass of MgCl₂** = 95 g/mol (Mg = 24 g/mol + 2 × Cl = 35 g/mol × 2)

\[ \text{Moles of NaCl} = \frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} = 0.1 \, \text{mol} \]
\[ \text{Moles of MgCl}_2 = \frac{9.50 \, \text{g}}{95 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 3: Calculate the effective number of moles after ionization
Now, we will consider the ionization of the solutes.

- For NaCl (80% ionization):
\[ \text{Moles of Na}^+ = 0.1 \, \text{mol} \times 0.80 = 0.08 \, \text{mol} \]
\[ \text{Moles of Cl}^- = 0.1 \, \text{mol} \times 0.80 = 0.08 \, \text{mol} \]

- For MgCl₂ (50% ionization):
\[ \text{Moles of Mg}^{2+} = 0.1 \, \text{mol} \times 0.50 = 0.05 \, \text{mol} \]
\[ \text{Moles of Cl}^- = 0.1 \, \text{mol} \times 0.50 \times 2 = 0.10 \, \text{mol} \]

### Step 4: Calculate total moles of particles in solution
Now we can find the total number of moles of particles in the solution:

\[ \text{Total moles} = \text{Moles of Na}^+ + \text{Moles of Cl}^- + \text{Moles of Mg}^{2+} + \text{Moles of Cl}^- \]
\[ = 0.08 + 0.08 + 0.05 + 0.10 = 0.31 \, \text{mol} \]

### Step 5: Calculate molality of the solution
To find molality (m), we need to convert the weight of the solvent from grams to kilograms:

\[ \text{Weight of solvent} = 84.65 \, \text{g} = 0.08465 \, \text{kg} \]
\[ \text{Molality (m)} = \frac{\text{Total moles}}{\text{Weight of solvent in kg}} = \frac{0.31 \, \text{mol}}{0.08465 \, \text{kg}} \approx 3.66 \, \text{mol/kg} \]

### Step 6: Calculate the boiling point elevation (ΔT_b)
Using the formula for boiling point elevation:

\[ \Delta T_b = K_b \times m \]
Where \( K_b \) for water is 0.51 °C kg/mol.

\[ \Delta T_b = 0.51 \, \text{°C kg/mol} \times 3.66 \, \text{mol/kg} \approx 1.87 \, \text{°C} \]

### Step 7: Calculate the normal boiling point of the solution
The normal boiling point of pure water is 100 °C. Therefore, the boiling point of the seawater solution is:

\[ \text{Boiling point} = 100 \, \text{°C} + \Delta T_b = 100 \, \text{°C} + 1.87 \, \text{°C} \approx 101.87 \, \text{°C} \]

### Final Answer
The normal boiling point of the seawater solution is approximately **101.87 °C**.

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